JEE Main 2026 — Hydrocarbons Question with Solution

The starting compound is 1-ethyl-4-nitrobenzene.

Reaction with ${\text{Br}}_2/{\text{AlBr}}_3$ is an electrophilic aromatic substitution. The $-{\text{C}}_2{\text{H}}_5$ group is activating and ortho/para-directing, while the $-{\text{NO}}_2$ group is deactivating and meta-directing. Both groups direct the incoming electrophile (${\text{Br}}^{+}$) to the position ortho to the $-{\text{C}}_2{\text{H}}_5$ group. Thus, compound A is 2-bromo-1-ethyl-4-nitrobenzene.

Reaction with $\text{Sn}/\text{HCl}$ reduces the $-{\text{NO}}_2$ group to an $-{\text{NH}}_2$ group, forming compound B (4-amino-2-bromo-1-ethylbenzene).

Reaction with ${\text{NaNO}}_2/\text{HCl}$ at $273-278\text{ K}$ converts the $-{\text{NH}}_2$ group into a diazonium salt, forming compound C (3-bromo-4-ethylbenzenediazonium chloride).

Reaction with ethanol (${\text{C}}_2{\text{H}}_5\text{OH}$) reduces the diazonium group, replacing it with a hydrogen atom. This forms compound D (1-bromo-2-ethylbenzene).

Reaction with alkaline ${\text{KMnO}}_4$ followed by acidic hydrolysis oxidizes the ethyl side chain ($-{\text{C}}_2{\text{H}}_5$) to a carboxylic acid group ($-\text{COOH}$). The bromine atom remains unaffected. This forms compound E (2-bromobenzoic acid).

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